SOLUTION: A normal population has a mean of 60 and a standard deviation of 3. You select a sample of 36.
Compute the probability the sample mean is (Round z values to 2 decimal places a
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-> SOLUTION: A normal population has a mean of 60 and a standard deviation of 3. You select a sample of 36.
Compute the probability the sample mean is (Round z values to 2 decimal places a
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You can put this solution on YOUR website! These all make use of finding the z value for each number x where z=(x-mean)/std dev.
The standard deviation of a sample mean is called the standard error of the mean, and it is sigma/sqrt(n), here 3/6 or 0.5. All the z s are integers, so I didn't write for example -2.00
for mean < 59 it is (59-60)/0.5=-2 so it is z < -2 or 0.0228
for mean between 59 and 61 it is z between -2 and +2 or a probability of 0.9545. 59 is 2 sd s to the left of the mean and 61 2 sd s to the right of the mean.
for mean between 61 and 62 it is 2 < z < 3 or 0.0214
for mean greater than 62 it is z>4 or 0.00003
